Statement
Consider the 2 well-known properties of a linear operator \(T : \mathbb{R}^m \to \mathbb{R}^n : m, n \in \mathbb{N}\),
\[\begin{align} T \left( \sum_a \pmb{u}_a \right) & = \sum_a T \left( \pmb{u}_a \right) & \forall \: \pmb{u}_a \in \mathbb{R}^m && \left( 1 \right) \\ T \left( c \pmb{u} \right) & = c T \left( \pmb{u} \right) & \forall \: c \in \mathbb{R}, \pmb{u} \in \mathbb{R}^m && \left(2 \right) \end{align}\]Proposition: Property \(\left( 2 \right)\) (homogeneity) can be derived from property \(\left( 1 \right)\) (additivity) using the underlying structure of \(\mathbb{R}\).
Thus, we propose that linearity can be unambiguously defined using axiom \(\left( 1 \right)\) only, at least up to \(\mathbb{R}^m\).
Overview of proof: Represent an arbitrary \(c \in \mathbb{R}\) as an infinite series of rationals, using Dedekind cuts. Plug the result into \(\left( 1 \right)\) and manipulate to obtain \(\left( 2 \right)\).
Let us now elaborate on the above.
Reals as infinite series of rationals
Suppose we are given any \(c \in \mathbb{R}\). The construction of \(\mathbb{R}\) from \(\mathbb{Q}\) using Dedekind cuts guarantees the following:
\[\forall \: c \in \mathbb{R} : \exists \: \left\{ a_n : n \in \mathbb{N} \right\} : c = \lim_{n \to \infty} a_n, a_n \in \mathbb{Q} \: \forall \: n \in \mathbb{N}\]Now, we can define another sequence of rationals \(\displaystyle{ \left\{ b_k : k \in \mathbb{N} \right\} : \sum_{k=1}^n b_k = a_n }\). It follows,
\[\begin{align} b_n & = \sum_{k=1}^n b_k - \sum_{k=1}^{n-1} b_k \\ & = a_n - a_{n-1} \\ c & = \lim_{n \to \infty} a_n \\ & = \lim_{n \to \infty} \sum_{k=1}^n b_k \\ & = \sum_{k=1}^\infty b_k \end{align}\]Thus, every real number can be represented as an infinite series of rationals,
\[\implies \forall \: c \in \mathbb{R} : \exists \: \left\{ b_k : k \in \mathbb{N} \right\} : c = \sum_{k=1}^\infty b_k, b_k \in \mathbb{Q} \: \forall \: k \in \mathbb{N}\]Furthermore, by definition, every rational is some integer divided by some non-zero integer,
\[\forall \: b \in \mathbb{Q} : \exists \: p \in \mathbb{Z}, q \in \mathbb{Z} \backslash \left\{ 0 \right\} : b = \frac{p}{q}\]Let us proceed to derive homogeneity from additivity for the special case \(c \in \mathbb{Q}\) (however, we will still have \(\pmb{u} \in \mathbb{R}^m\)). Then, we will apply the result above to extend our observations to \(c \in \mathbb{R}\).
Suppose \(c \in \mathbb{Q} , \pmb{u} \in \mathbb{R}^m\). Consider a linear operator \(T : \mathbb{R}^m \to \mathbb{R}^n\) which obeys additivity i.e. property \(\left( 1 \right)\). Let us express \(c\) as \(c = \frac{p}{q} : p \in \mathbb{Z}, q \in \mathbb{Z} \backslash \left\{ 0 \right\}\). The signature of \(c\) can be encoded in either the numerator \(p\), which, pedantically speaking, is preferred over \(q\) as the domain of the former is the entirety of \(\mathbb{Z}\).
For \(c \in \mathbb{Z}\)
We will first derive homogeneity from additivity for the simple case \(q=1, p \in \mathbb{Z} \implies c \in \mathbb{Z}\). To do so, we will investigate the individual scenarios \(p \in \mathbb{Z}^+, p=0, p \in \mathbb{Z}^-\).
For \(c \in \mathbb{Z}^+ = \mathbb{N}\)
We have,
\[\begin{align} T \left( c \pmb{u} \right) & = T \left( p \pmb{u} \right) : p \in \mathbb{N} \\ & = T \left( \sum_{a = 1}^p u \right) \end{align}\]By additivity,
\[\begin{align} T \left( c \pmb{u} \right) & = \sum_{a=1}^p T \left( \pmb{u} \right) \\ & = p T \left( \pmb{u} \right) \\ & = c T \left( \pmb{u} \right) & \square \end{align}\]For \(c=0\)
For \(c=0\), we have,
\[\begin{align} T \left( 0 \cdot \pmb{u} \right) & = T \left( \pmb{0}_m \right) \\ & = T \left( \pmb{0}_m + \pmb{0}_m \right) \\ & = T \left( \pmb{0}_m \right) + T \left( \pmb{0}_m \right) \\ T \left( \pmb{0}_m \right) & = T \left( \pmb{0}_m \right) + T \left( \pmb{0}_m \right) \\ \implies T \left( \pmb{0}_m \right) & = \pmb{0}_n \end{align}\]where \(\pmb{0}_k\) is the null vector in \(\mathbb{R}^k\).
For \(c \in \mathbb{Z}^-\)
Let \(p = -n : n \in \mathbb{Z}^+\),
\[\begin{align} \pmb{0}_n & = T \left( \pmb{0}_m \right) \\ & = T \left( n \pmb{u} - n \pmb{u} \right) \end{align}\]By additivity,
\[\begin{align} \pmb{0}_n & = T \left( n \pmb{u} \right) + T \left( -n \pmb{u} \right) \\ & = n T \left( \pmb{u} \right) + T \left( -n \pmb{u} \right) & \left[ \because n \in \mathbb{Z}^+ \right] \\ \implies T \left( p \pmb{u} \right) & = T \left( - n \pmb{u} \right) \\ & = \pmb{0}_n - n T \left( \pmb{u} \right) \\ & = - n T \left( \pmb{u} \right) \\ & = p T \left( \pmb{u} \right) & \square \end{align}\]Combining the different scenarios above, we have indeed found,
\[\begin{align} \forall \: c \in \mathbb{Z} : \left( 1 \right) \implies \left( 2 \right) && \left( A \right) \end{align}\]For \(c \in \left\{ \frac{1}{q} : q \in \mathbb{Z} \backslash \left\{ 0 \right\} \right\}\)
We will now see how the above result is also true for \(c = \frac{p}{q} : p = 1, q \in \mathbb{Z} \backslash \left\{ 0 \right\}\). Ultimately, we will combine this result with the corresponding one for \(c \in \mathbb{Z}\) to generalize it for all \(c \in \mathbb{Q}\) which in turn will let us generalize it to \(c \in \mathbb{R}\) via Dedekind cuts as previously stated.
We begin with the statement,
\[\begin{align} T \left( \sum_{a=1}^q \frac{1}{q} \pmb{u} \right) & = T \left( \pmb{u} \right) \\ \sum_{a=1}^q T \left( \frac{1}{q} \pmb{u} \right) & = T \left( \pmb{u} \right) \\ q T \left( \frac{1}{q} \pmb{u} \right) & = T \left( u \right) \\ T \left( \frac{1}{q} \pmb{u} \right) & = \frac{1}{q} T \left( u \right) & \square \end{align}\]Thus,
\[\begin{align} \forall \: c = \frac{1}{q} : q \in \mathbb{Z} \backslash \left\{ 0 \right\} : \left( 1 \right) \implies \left( 2 \right) && \left( B \right) \end{align}\]For \(c \in \mathbb{Q}\)
Let \(c = \frac{p}{q} : p \in \mathbb{Z}, q \in \mathbb{Z} \backslash \left\{ 0 \right\}\). Using the previous results \(\left( A \right)\) and \(\left( B \right)\),
\[\begin{align} T \left( c \pmb{u} \right) & = T \left( \frac{p}{q} \pmb{u} \right) \\ & = T \left( p \cdot \frac{1}{q} \pmb{u} \right) \\ & = p T \left( \frac{1}{q} \pmb{u} \right) & \left[ \left( A \right) \right] \\ & = \frac{p}{q} T \left( \pmb{u} \right) & \left[ \left( B \right) \right] \\ & = c T \left( \pmb{u} \right) & \square \end{align}\] \[\begin{align} \forall \: c \in \mathbb{Q} : \left( 1 \right) \implies \left( 2 \right) && \left( C \right) \end{align}\]For \(c \in \mathbb{R}\)
Consider an arbitrary \(c \in \mathbb{R}\). From the section on reals as infinite series of rationals, there exists a sequence of rationals \(\left\{ b_k : k \in \mathbb{N} \right\} : b_k \in \mathbb{Q} \: \forall \: k \in \mathbb{N}\) such that it adds up to \(c\),
\[\displaystyle{ c = \sum_{k=1}^\infty b_k }\]Hence,
\[T \left( c \pmb{u} \right) = T \left( \sum_{k=1}^\infty b_k \pmb{u} \right)\]By additivity,
\[\begin{align} T \left( c \pmb{u} \right) & = \sum_{k=1}^\infty T \left( b_k \pmb{u} \right) \\ & = \sum_{k=1}^\infty b_k T \left( \pmb{u} \right) & \left[ \left( C \right) \right] \\ & = c T \left( \pmb{u} \right) & \blacksquare \end{align}\] \[\forall \: c \in \mathbb{R} : \left( 1 \right) \implies \left( 2 \right)\]Summary
To summarize the above approach, homogeneity of linear operators on vector spaces built on the base field \(\mathbb{R}\), comes for free from their additivity. We realized this by using the construction of \(\mathbb{R}\) from \(\mathbb{Q}\) which is in turn constructed from \(\mathbb{Z}\).
The advantage of expressing reals in terms of integers is that integers can fundamentally be used for counting, which is implicitly applied in property \(\left( 1 \right)\), additivity. Thus, this bridge from \(\left( 1 \right)\) to \(\left( 2 \right)\) allows us to logically proceed in the same direction.